Our aim is to find a condition for the existence of a hyperbolic
circle
through the vertices of a hyperbolic triangle in terms of the lengths
of its sides.
We begin with a theorem which holds in both euclidean and hyperbolic
geometry, i.e. a theorem of neutral geometry.
In the theorem, we consider the size of angles, and assume that each
angle is taken in the range (0,π). That is to say, we take <XYZ as the
measure of the internal angle at vertex Y in ΔXYZ.
the angle theorem
For any triangle ΔABC, and any point P which is not on any
of the lines AB, BC, CA. Let <APB=γ, <BPC=α, <CPA=β, then
cos2(α) + cos2(β) + cos2(γ) =
2cos(α)cos(β)cos(γ) + 1.
proof
Although this result has nice applications in euclidean geometry, we
shall
concentrate on the hyperbolic case.
basic lemma
Suppose that ΔABC is a hyperbolic triangle
and that P is a point
not on any of the hyperbolic lines AB, BC, CA.
Let d(A,B)=c, d(B,C)=a, d(C,A)=b, d(P,A)=r, d(P,B)=s, d(P,C)=t.
Then G(cosh(r),cosh(s),cosh(t),cosh(a),cosh(b),cosh(c)) = 0,
where G(R,S,T,X,Y,Z) is given by
(T2-1)(RS-Z)2 +
(S2-1)(RT-Y)2 +
(R2-1)(TS-X)2
-2(RS-Z)(ST-X)(TR-Y) - (R2-1)(S2-1)(T2-1).
This is obtained from the angle theorem and the cosine rule
for the angles at P.
It is easy to see that G is quadratic in each of R, S, T.
A calculation shows that the coefficient of T2 is Z2-1.
Since we will have Z = cosh(c) > 1, G is quadratic in T.
Now observe that, if ΔABC has a hyperbolic circumcircle K(P,r), then
we can apply the lemma to the hyperbolic centre P. We then have
R = S = T = cosh(r), and the condition becomes
(R2-1){(R2-Z)2+
(R2-Y)2+
(R2-X)2}-2(R2-Z)(R2-X)(R2-Y)-
(R2-1)3 = 0,
where X =cosh(a), Y = cosh(b), Z = cosh(c), R = cosh(r).
The left hand side is clearly a polynomial in R2, and appears to involve
terms in R4 and R6. If we multiply out and simplify, we get the equation
F(cosh(a),cosh(b),cosh(c))R2 = Δ2(cosh(a),cosh(b),cosh(c)),
where
F(X,Y,Z) = 3+2(XY+YZ+ZX)-2(X+Y+Z)-(X2+Y2+Z2), and
Δ2(X,Y,Z) = 1+2XYZ-(X2+Y2+Z2).
This is the the circumradius equation
By expanding then simplifying we get the difference formula
Δ2(X,Y,Z)-F(X,Y,Z) = 2(X-1)Y-1)(Z-1).
We have met Δ2 before, in connection with the trigonometry of hyperbolic
triangles. From that context, we know that it must be positive. We take
Δ
as the positive root of this quantity.
It follows that, as R2 > 0, we must have F(cosh(a),cosh(b),cosh(c)) > 0.
We have arrived at the following
necessary condition for a hyperbolic circumcircle
If the hyperbolic triangle ΔABC has a hyperbolic circumcircle, then
(1) F(cosh(a),cosh(b),cosh(c)) > 0, and
(2) the hyperbolic radius r is determined by
cosh(r) = Δ(cosh(a),cosh(b),cosh(c))/F½(cosh(a),cosh(b),cosh(c)).
The other hyperbolic functions of r are easily determined from cosh(r).
The manipulation is a little tedious, but we obtain
sinh2(r) = 2(cosh(a)-1)cosh(b)-1)(cosh(c)-1)/F(cosh(a),cosh(b),cosh(c)),
tanh2(r) = 2(cosh(a)-1)cosh(b)-1)(cosh(c)-1)/Δ2(cosh(a),cosh(b),cosh(c)).
The formula for sinh(r) shows again that we must have F > 0.
The formula for tanh(r) gives an alternative formulation of the condition
Alternative necessary condition
If ΔABC has a hyperbolic circumcircle, then we must have
2(cosh(a)-1)cosh(b)-1)(cosh(c)-1) < Δ2(cosh(a),cosh(b),cosh(c)).
This is immediate as tanh(r) < 1.
If we observe that cosh(x)-1 = 2sinh2(½x), the formula for tanh(r) yields
tanh(r) = 4sinh(½a)sinh(½b)sinh(½c)/Δ(cosh(a),cosh(b),cosh(c))
The removal of the need to take a root is an illusion - Δ involves a root.
There are various ways to rewrite the formulae for r. Some are given as
a digression.
Our main purpose is to show that either of our conditions are also sufficient
to guarantee the existence of a circumcircle. This is not so easy. We give a
proof in a separate page.
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